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Chapter 10: Gene Mutation: Origins and Repair Processes Multiple-Choice Questions 1. Newcombe spread E. coli cells on an agar base. After several generations of growth, he respread the cells and sprayed them with streptomycin, thus killing all cells except those that were resistant mutants. More mutants were observed after spreading than if they had not been respread. The experiment was used to A) show how to screen for environmental mutagens. B) show that mutation occurs in prokaryotes as well as in eukaryotes. C) demonstrate that in some cases mutations are caused by the selective agent itself. * D) provide evidence that mutations occur in the absence of the selective agent. E) show a direct correlation between the amount of the selective agent used and the number of resistant mutants (one-hit relationship). 2. In E. coli a region of a gene with repeats of the sequence CTGG will be prone to A) reversion. B) missense mutation. C) nonsense mutation. * D) frameshift mutation. E) amber mutation. 3. In E. coli a region flanked by two repeats of a sequence such as GTGGTGAA is prone to * A) deletion. B) missense mutation. C) duplication. D) inversion. E) frameshift mutation. 4. Fragile X syndrome is caused by * A) trinucleotide repeats. B) free radicals. C) microdeletions. D) 5-Bromouracil. E) depurination. 5. During mutagenic treatment with nitrous acid, an adenine deaminates to form hypoxanthine, which bonds like guanine. The mutational event would be A) AT to CG * B) AT to GC C) AT to TA D) GC to AT E) GC to TA 6. In E. coli, mutation arising during repair is mostly by A) thymine dimer splitting. B) excision repair. C) mismatch repair. D) recombinational repair. * E) SOS repair. Test Bank: Chapter Ten 7. The fluctuation test of Luria and Delbruck established that A) T1 phage was probably not a mutagen. B) mutations could arise prior to the time they were selected. C) the mutation rate was constant per cell per generation in constant conditions. D) in E. coli the number of mutants per clone was relatively constant. E) All the above * F) A, B, C. G) A, B, D. 8. Trinucleotide repeat disorders are the result of A) a high rate of mutation throughout the genome. * B) extensive duplication of a single codon. C) deviations from the genetic code in human mitochondria. D) tRNAs failing to recognize specific codons. E) transversions of the DNA bases in a coding sequence. 9. If an incorrect base is incorporated during DNA synthesis and is not corrected by DNA polymerase, it can be corrected by postreplication repair. This involves A) detection of the mismatch. B) a recognition of the methylation status of the DNA strands. C) a process similar to excision repair. * D) All the above. E) None of the above. 10. In the galactose operon, the order of genes is promoter–epimerase–transferase– galactokinase. An IS element in the galactokinase gene will knock out function * A) only in galactokinase. B) only in epimerase. C) only in transferase. D) in galactokinase, transferase, and epimerase. E) in galactokinase and transferase. 11. Hybridization of single-stranded wild-type DNA with DNA from mutations caused by IS elements characteristically shows (through electron microscopy) A) chi structures. B) unpaired tails. * C) a single-stranded loop representing IS DNA. D) a single-stranded loop representing wild-type DNA. E) theta structures. 12. Retroviruses replicate using A) DNA polymerase. B) RNA polymerase. C) restriction endonuclease. * D) reverse transcriptase. E) topoisomerase. Gene Mutation: Origins and Repair Processes 13. Transposons are useful in genetic analysis because of their use in * A) gene tagging. B) preventing crossing-over. C) stabilizing aneuploids. D) generating duplications. E) preventing reversion of mutant alleles. 14. Retrotransposons move via an intermediate that is A) a double-stranded lollipop. B) a retrovirus. C) double-stranded RNA. D) single-stranded DNA. * E) single-stranded RNA. 15. Bacterial transposon structure can be thought of as A) IS sequences flanked by inverted drug-resistance genes. * B) drug-resistance gene(s) flanked by IS elements. C) drug-resistance gene(s) flanked by a pair of mu phages. D) a mu phage flanked by two IS elements. E) resistance gene(s) flanked by inverted resistance transfer factors (RTFs). 16. A bacterial cell contains a single copy of a transposon in which one strand carries a G at one site (the wild-type base at this site) and the other strand carries a T (which confers a mutant phenotype). If this cell is allowed to divide through several divisions, but the transposon transposes at the first division through replicative transposition, then the colonies derived from these cells will be A) mosaic, with wild-type and mutant sectors. B) all mutant. C) all wild-type. D) all GT heteroduplexes. * E) some wild and some mutant. 17. A corn plant is homozygous for a mutant allele that results in no color in the seed (white). The mutant is caused by Ds insertion that often exits late in seed development, and there is an active Ac element in the genome. The seeds will be A) white. B) pigmented all over. * C) white with small spots of pigment. D) white with large spots of pigment. E) weakly pigmented. Test Bank: Chapter Ten Definitions Give the word or phrase that best defines each statement below. 1. Type of mutation in which a pyrimidine is substituted for a purine Answer: Transversion 2. Rare state of a normal base which may lead to mutation by its faulty base-pairing properties Answer: Tautomer (enol) 3. Ultraviolet dark-repair system least prone to formation of mutations Answer: Excision repair 4. A postreplication repair system most prone to error Answer: Recombinational repair 5. Type of mutation in which a purine is substituted for a purine Answer: Transition 6. Mutagen causing base changes from GC to AT only Answer: Hydroxylamine 7. Mutation caused by acridine Answer: Frameshift 8. A base change resulting in a codon specifying the same amino acid Answer: Silent mutation 9. A base change resulting in a codon specifying a different amino acid Answer: Missense mutation 10. A base change resulting in a stop codon Answer: Nonsense mutation 11. An enzyme that repairs thymine dimers in visible light in E. coli Answer: Photolyase 12. An enzyme that attacks sites left after loss of a single purine Answer: AP endonuclease 13. A mutation that causes a mutant phenotype only under restrictive conditions Answer: Conditional mutation Gene Mutation: Origins and Repair Processes Open-Ended Questions 1. A sample of food from a University of Michigan dining hall was found by a genetics student to induce mutation in E. coli in high frequency. The single active ingredient was found to be the chemical saltpeter. Several arg mutants were obtained after treatment with the compound. The abilities of known mutagens to reverse the saltpeter mutations (to arg+) were determined with a large number of the induced mutants. The frequency of arg cells among total cells after treatment was: No treatment (control) 6.8 × 106 2-Aminopurine 8.0 × 104 Saltpeter 5.8 × 106 Acridine 7.0 × 106 5-Bromouracil 6.1 × 104 Hydroxylamine 6.9 × 106 What DNA base changes are caused by dining hall food laced with saltpeter? Explain your reasoning. Answer: GC AT; saltpeter appears to work like hydroxylamine which cannot bring about reversion of a GC AT transition. 5-Bromouracil and 2-aminopurine can. Acridine results in frameshift mutations and, thus, cannot revert transitions or transversions. 2. Three ara mutants of E. coli were induced by mutagen X. The ability of other mutagens to cause the reverse change (ara ara+) was tested, and the frequency of ara+ cells among total cells after treatment is shown below: Mutant ara1 ara2 ara3 None 1.5 × 108 2.0 × 107 6.0 × 107 Mutagen tested 5-Bromouracil 2-Aminopurine 5.0 × 105 1.8 × 104 4 2.0 × 10 6.1 × 105 1.1 × 105 9.0 × 106 Hydroxylamine Proflavin 1.8 × 108 1.6 × 108 5 3.0 × 10 1.6 × 107 8.2 × 106 6.6 × 107 Assume all ara cells are true revertants. What base changes were probably involved in forming the three original mutations (ara+ ara)? What kind or kinds of mutations are caused by mutagen X? (a) ara+ ara1: (b) ara+ ara2: (c) ara+ ara3: Answer: (a) GC AT: reverted by base analogs but not by hydroxylamine or proflavin. (b) AT GC: reverted by base analogs and hydroxylamine but not by proflavin. (c) AT GC: see (b); mutagen X causes transition mutations in both directions, since mutants are revertible by base analogs, some are revertible by hydroxylamine, and none (if this is a representative sample) are revertible by proflavin. Test Bank: Chapter Ten 3. (a) A met+ strain of Neurospora was treated with a mutagen to create met- mutants. The mutants were reverted to met+ with HA (hydroxylamine). What was the original mutation on the molecular level? Name a possible mutagen that could have been used to cause the original mutation. (b) Two of the revertants showed odd results when crossed with a wild-type met+ strain. met+ revertant A x met+ wild-type 88% met+ 12% metmet+ revertant B x met+ wild-type 93% met+ 7% metExplain how these results occurred and why the above numbers of offspring were obtained. Answer: (a) HA causes GC to AT transitions so the original mutation must have been AT to GC. Possibly caused by nitrous acid. (b) This reversion is not true reversion but suppression. Furthermore, in case A the suppressor must be 24 m.u. away from the met+ locus, although in B it is 14 m.u. away. 4. Show the steps (i.e., several replications) by which 5-bromouracil causes the GC base pair shown below to change to an AT base pair. Show the point at which the rare (tautomeric) form of 5-bromouracil occurs. Show both products of all divisions in which 5-bromouracil is involved. Do not show chemical structures. Answer: Gene Mutation: Origins and Repair Processes 5. A bacterium E. coli suffered ultraviolet irradiation. In the base-pair sequence shown below, note that the adjacent thymines formed a dimer. What are the requirements, the process, and the outcome of the repair by (a) photoreactivation and (b) general excision repair, if this damage were to be repaired? —————————————————— A G C T = T A G C T C G A A T C G —————————————————— Answer: (a) (b) Requirements Photolyase, visible light uvrA, B, C (excinuclease); helicase II, DNA polymerase I; no visible light (i.e., dark) Process Reversal of dimerization uvrA recognizes damaged DNA, complexes with uvrB;uvrC binds to B and each makes incision; 12-mer unwound and released by helicase II; DNA pol I repair synthesis and ligase fill in missing patch. Outcome You end up with the same Replacement of bases using complementary DNA (same chains present) strand. as before UV irradiation; same bases present; no repair synthesis required. 6. PCR (polymerase chain reaction) was used to amplify a small region at the 5´ end of the coding region of three mutant alleles in yeast. This region was sequenced as follows, grouped in codons: Wild type 5´ GAA CTC GAG CTT AAT 3´ Mutant 1 5´ GAA CTC GAG CTT ATT 3´ Mutant 2 5´ GAA CTC AAG CTT AAT 3´ Mutant 3 5´ GAA CTC GAG CCT TAA T 3´ Classify these mutations into as many categories as possible. (a) Mutant 1 (b) Mutant 2 (c) Mutant 3 Answer: (a) Mutant 1 shows A to T, so it is an AT to TA transversion, and missense. (b) Mutant 2 shows G to A, so it is a GC to AT transition, and also missense. (c) Mutant 3 is a frameshift mutation caused by an insertion of a C; TAA is a stop codon making this a nonsense mutation. 7. Three mutations were obtained in a bacterial gene for whose protein product an antibody is available. Both Northern analysis (RNA separated by electrophoresis, blotted, and probed with DNA) and Western analysis (proteins separated by electrophoresis, blotted, and probed with antibodies) were performed on the mutants. The results are summarized below: Test Bank: Chapter Ten For each mutation, what kind of mutation occurred and how do you know it? (a) Mutant 1 (b) Mutant 2 (c) Mutant 3 Answer: (a) Mutant 1 is a missense or frameshift mutation (no change in RNA or protein size). (b) Mutant 2 is a nonsense mutation (RNA same size but protein shorter). (c) Mutant 3 is possibly a promoter mutation as no RNA is transcribed, or possibly a deletion of the gene. 8. Assume that 5-bromouracil (BU), kept during one cycle of DNA replication only, causes the mutation GC AT. (a) Draw in complete detail the “abnormal” base pair that is ultimately responsible for the mutation. Show only the purine and pyrimidine “half-rings” that contain the H-bonding groups. (b) Draw the sequence of events and replication cycles by which the mutant sequence is generated. Start with a double-stranded segment having the sequence 5´-CAT-3´ in one strand. (c) Is the mutation a transition or a transversion? Answer: (a) (b) Gene Mutation: Origins and Repair Processes Note: The new strand has been shown in lowercase letters at each replication step. Wildtype molecules were not replicated further. (c) Transition 9. Starting with the segment of a double-stranded DNA molecule shown, give the sequence of the molecule resulting from each of the mutations indicated below. Show 3´ and 5´ ends, and write mutant base pairs in lowercase letters. (Position numbers below refer to the top strand.) 5´-AACCTTGGAA-3´ 3´-TTGGAACCTT-5´ (a) A transition at base pair position 5 (b) A transversion at base pair position 5 (c) A deletion of two base pairs at positions 4 and 5, respectively (d) An insertion of three base pairs between base pairs 5 and 6 (e) An inversion of the three base pairs beginning at position 3 Answer: (Bases changed by mutation shown in small letters) 5´-AACCTTGGAA-3´ 3´-TTGGAACCTT-5´ Wild-type 5´-AACTGGAA-3´ 3´-TTGACCTT-5´ (c) 5´-AACCcTGGAA-3´ 3´-TTGGgACCTT-5´ (a) 5´-AACCTtttTGGAA-3´ 3´-TTGGAaaaACCTT-5´ (d) 5´-AACCaTGGAA-3´ 3´-TTGGtACCTT-5´ (b) 5´-AAaggTGGAA-3´ 3´-TTtccACCTT-5´ (e) Note: 5´-AAtccTGGAA-3´ 3´-TTaggACCTT-5´ is not a correct answer for (e) because this would create a segment of 3´-to-5´ polarity within the main 5´-to-3´ polarity of each strand. 10. A corn plant carries mutant alleles for the single gene that deposits colored pigment in leaves. (When active, this gene deposits purple pigment that, with the green of chlorophyll, makes the plant brown.) The plant in question is green with brown spots all over it. When the plant is selfed, 1/4 of the progeny are plain green and 3/4 are green with brown spots. (a) Explain the spotted phenotype of the original plant. (b) Explain the 3:1 ratio. (c) Could you ever obtain a fully brown progeny from this self-cross? Answer: (a) An active transposon has entered the gene that causes pigment deposition, thereby inactivating it. Frequent excisions cause pigmented spots, one from each excision. (b) The plant is heterozygous in some way. Perhaps it is aua, where au is an unstable allele caused by transposon insertion and a is a stable allele caused by a base pair change. In that case, 3/4 of the progeny will be au and spotted. (c) Yes, if a reverted, or if the transposon excised in the germinal tissue. Test Bank: Chapter Ten 11. A yeast strain of wild-type phenotype carried a single copy of an active transposon. This strain was plated, and one rare colony on the plate was found to be extremely small; subsequent studies showed that it had very thin cell walls. A Southern analysis revealed a new fragment hybridizing to the transposon probe, showing that the small strain now had two copies of the transposon. The new fragment was extracted and, using the known transposon sequence, a primer was designed to sequence outward from the transposon tip. The sequence obtained was part transposon DNA, but when the next 500 nucleotides were tested against the DNA data bank, its putative amino acid sequence was found to be glucose-6-phosphate dehydrogenase. Explain how the mutant phenotype arose, and the significance of the G6PD. Answer: From its original benign location (presumably not inside a gene), a copy of the transposon has moved into the G6PD gene, inactivating it. The sequence is a junction between the transposon and the gene. G6PD is normally active in carbohydrate metabolism; when inactive through mutation, it can apparently interfere with cell wall synthesis (polysaccharide synthesis). 12. A gene G, containing two small introns, was inserted into a yeast transposon at a position where it did not interfere with transposition. The transposon moved to new locations, and the structure of the transposon in its new locations was studied. It was found that at each new location the transposon still contained gene G but without introns, in every case. Explain the loss of the introns, and state what this tells us about the mechanism of transposition. Answer: The transposon must be a retrotransposon, which moves via an RNA intermediate. The introns must have been spliced out of the RNA, and then reverse transcriptase made DNA copies, which were inserted into the new locations. 13. In Neurospora, the cloned am (amination) gene hybridizes to a single 4-kb HaeIII genomic fragment in a wild-type strain from Africa. Three am mutants arose spontaneously in this strain. The probe was used on HaeIII digests of these three strains. In two of them the probe bound to a single 9-kb fragment, and in the third one the probe bound to two fragments of 3 kb and 1 kb. What is likely to be the nature of these three mutants? Answer: The types with 9-kb fragments are probably caused by the insertion of a 5-kb transposon into the am gene. The other mutation is probably a spontaneous missense mutation that happened to create an HaeIII site within the am gene. 14. (a) What elements must a transposon contain? (b) What are minimum elements that a mutant transposable element must have in order to be transposed in a cell that harbors a separate, complete transposon? Answer: (a) Genes for transposase and terminal inverted repeats that the transposase recognizes (b) Terminal inverted repeats that the complete transposon’s transposase recognizes Gene Mutation: Origins and Repair Processes 15. Explain the difference between the replicative and conservative modes of transposition. Answer: In replicative transposition, a new copy of the transposable element is generated during transposition. In conservative transposition, the transposon is not replicated. The element is excised from its location and integrated into a new site. 15. A transposon was found as an insertion in the waxy gene of corn. A portion of the DNA sequence of this transposon and the waxy gene is given below: This transposon is known to possess 9-kb inverted repeats and to produce a 6-bp target site duplication. (a) Draw a line under the 9-bp inverted repeat, and draw a box around the 6-bp target site duplication. (b) Fill in the 20 missing nucleotides of DNA sequence at the right end of the transposon insertion. (Place an X in any position where it is not possible to determine the specific nucleotide.) Answer: (a) (b) Test Bank: Chapter Ten 17. The C gene of corn determines solid colored pigment. A corn plant of genotype c/cDs ; Ac/Ac+ produces spotted kernels; cDs is an unstable allele caused by insertion of a Ds element. The symbol Ac+ indicates a chromosome that does not possess an Ac element. A cross is made between this plant and a colorless corn plant of genotype c/c ; Ac+/Ac+. What are the expected genotypes and phenotypes, and their proportions among the progeny from this cross? Assume that Ac and c are unlinked. Answer: c/c; Ac/A + 1 / 4 colorless c/c; Ac + /Ac + 1 / 4 colorless 3 / 4 colorless c Ds /c; Ac+ /Ac + 1 / 4 colorless c Ds /c; Ac/Ac + 1 / 4 spotted 1 / 4 spotted 18. Transposons have been referred to as “jumping genes” because they appear to jump from one position to another, leaving the old locus and appearing at a new locus. (a) What are the two general mechanisms of transposition in bacteria? (b) Which of these mechanisms applies to the concept of “jumping genes”? Answer: (a) In bacteria there are two modes of transposition: (1) Conservative mode in which the transposon excises from its original position and inserts at a new position, and (2) Replicative mode in which the transposon moves to a new location by replicating into the target DNA, leaving behind a copy of the transposon at the original site. (b) The conservative mode results in true jumping genes.